Substring

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 485 Accepted Submission(s): 202

Problem Description

?? is practicing his program skill, and now he is given a string, he has to calculate the total number of its distinct substrings.
But ?? thinks that is too easy, he wants to make this problem more interesting.
?? likes a character X very much, so he wants to know the number of distinct substrings which contains at least one X.
However, ?? is unable to solve it, please help him.

Input

The first line of the input gives the number of test cases T;T test cases follow.
Each test case is consist of 2 lines:
First line is a character X, and second line is a string S.
X is a lowercase letter, and S contains lowercase letters(‘a’-‘z’) only.

T<=30
1<=|S|<=10^5
The sum of |S| in all the test cases is no more than 700,000.

Output

For each test case, output one line containing “Case #x: y”(without quotes), where x is the test case number(starting from 1) and y is the answer you get for that case.

Sample Input

2 a abc b bbb

Sample Output

Case #1: 3 Case #2: 3

Hint

In first case, all distinct substrings containing at least one a: a, ab, abc. In second case, all distinct substrings containing at least one b: b, bb, bbb.

题意：给你一个字符和一个字符串，问你字符串中包含这个字符的不同子串的个数。

分析：求出所有后缀后边最近的要求字符的位置f[i]，对于后缀sa[i]，对答案的贡献就是n - max(sa[i]+height[i],f[sa[i]])。

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
//rank从0开始
//sa从1开始,因为最后一个字符(最小的)排在第0位
//height从2开始,因为表示的是sa[i-1]和sa[i]
const int MAXN = 100005;
int T,n,m,Rank[MAXN],sa[MAXN],X[MAXN],Y[MAXN],height[MAXN],s[MAXN],f[MAXN];
int buc[MAXN];
char str[MAXN],C[2];
void calheight(int n)
{int i , j , k = 0;for(i = 1 ; i <= n ; i++) Rank[sa[i]] = i;for(i = 0 ; i < n ; height[Rank[i++]] = k)for(k?k--:0 , j = sa[Rank[i]-1] ; s[i+k] == s[j+k] ; k++);
}
bool cmp(int *r,int a,int b,int l)
{return (r[a] == r[b] && r[a+l] == r[b+l]);
}
void suffix(int n,int m = 128)
{int i , l , p , *x = X , *y = Y;for(i = 0 ; i < m ; i ++) buc[i] = 0;for(i = 0 ; i < n ; i ++) buc[ x[i] = s[i]  ] ++;for(i = 1 ; i < m ; i ++) buc[i] += buc[i-1];for(i = n - 1; i >= 0 ; i --) sa[ --buc[ x[i] ]] = i;for(l = 1,p = 1 ; p < n ; m = p , l *= 2) {p = 0;for(i = n-l ; i < n ; i ++) y[p++] = i;for(i = 0 ; i < n ; i ++) if(sa[i] >= l) y[p++] = sa[i] - l;for(i = 0 ; i < m ; i ++) buc[i] = 0;for(i = 0 ; i < n ; i ++) buc[ x[y[i]] ] ++;for(i = 1 ; i < m ; i ++) buc[i] += buc[i-1];for(i = n - 1; i >= 0 ; i --) sa[ --buc[ x[y[i]] ] ] = y[i];for(swap(x,y) , x[sa[0]] = 0 , i = 1 , p = 1 ; i < n ; i ++)x[ sa[i] ] = cmp(y,sa[i-1],sa[i],l) ? p-1 : p++;}calheight(n-1);//后缀数组关键是求出height,所以求sa的时候顺便把rank和height求出来
}
int main()
{scanf("%d",&T);for(int t = 1;t <= T;t++){scanf("%s",C);scanf("%s",str);int n = strlen(str);for(int i = 0;i < n;i++) s[i] = str[i];s[n] = 0;int now = n;for(int i = n-1;i >= 0;i--){if(str[i] == C[0]) now = i;f[i] = now;}suffix(n+1,128);long long tot = 1ll*n - max(sa[1],f[sa[1]]);for(int i = 2;i <= n;i++) tot += 1ll*n - max(sa[i]+height[i],f[sa[i]]);printf("Case #%d: %I64d\n",t,tot);}
}

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